Question

An altitude of a triangle is five-third of the length of its corresponding base. If the altitude is increased by four centimeter and the base decreased by two centimeter, the area of the triangle remains the same. Find the base and altitude of the triangle.

Answer 1

Hint:- In this question consider the base of triangle be $x$, therefore altitude of the triangle be $\dfrac{5}{3}x$ and new altitude and base would be $\dfrac{5}{3}x + 4$ and $x - 2$ hence we will get two equations for areas and then equate them to solve for $x$.

Complete step-by-step answer:

Given that,
Altitude of triangle if $\dfrac{5}{3}$ of the length of its corresponding base
Let the base of triangle be $x$
Therefore altitude $ = \dfrac{5}{3}x$
As we know that area of triangle $ = \dfrac{1}{2} \times {\text{base}} \times {\text{altitude}}$
Therefore area of triangle $ = \dfrac{1}{2} \times x \times \dfrac{5}{3}x = \dfrac{{5{x^2}}}{6}$.....take this as equation first
Now altitude be increased by four centimeter and the base decreased by two centimeter
Therefore new altitude $ = \dfrac{5}{3}x + 4 = \dfrac{{5x + 12}}{3}$
new base $ = x - 2$
Therefore new area of triangle $ = \dfrac{1}{2} \times \left( {x - 2} \right) \times \left( {\dfrac{{5x + 12}}{3}} \right)$......take this as equation second
It is given in the question that the area of the triangle remains the same as altitude increases and base decreases.
Therefore equation first is equals to equation second:
$
  \therefore \dfrac{{5{x^2}}}{6} = \dfrac{1}{2} \times \left( {x - 2} \right) \times \left( {\dfrac{{5x + 12}}{3}} \right) \\
  \dfrac{{5{x^2}}}{6} = \dfrac{{\left( {x - 2} \right) \times \left( {5x + 12} \right)}}{6} \\
  5{x^2} = \left( {x - 2} \right) \times \left( {5x + 12} \right) \\
  5{x^2} = 5{x^2} + 12x - 10x - 24 \\
  0 = 2x - 24 \\
  2x = 24 \\
  x = 12 \\
$
Hence the base of triangle $ = 12{\text{cm}}$
Altitude of triangles $ = \dfrac{5}{3}x = \dfrac{5}{3} \times 12 = 20{\text{cm}}$

Note:- Firstly we found the initial area of triangle using formula one-half multiplied by base and altitude and took that as first equation of area after that we found the new area in which altitude increased by four centimeter and base decreased by two centimeter and took that as equation second and as given in question that area of triangle remains same so we formed the equation, simplified it and found the base and altitude of triangle.